CSR-DCF视频目标跟踪论文笔记（2）——关于滤波器Learning的推导（Augmented Lagrangian方法）

1. 论文基本信息

• 论文标题：Discriminative Correlation Filter with Channel and Spatial Reliability
• 作者：Alan Lukezic等
• 出处：CVPR，2017
• 文章链接：https://arxiv.org/abs/1611.08461
• 补充材料：https://www.semanticscholar.org/paper/Discriminative-Correlation-Filter-with-Channel-and-Luke%C5%BEi%C4%8D-Voj%C4%B1-%C5%99/7b485979c75b46d8c194868c0e70890f4a0f0ede
• 源码链接：https://github.com/alanlukezic/csr-dcf

这篇笔记主要针对滤波器求解的推导过程进行分析（拉格朗日乘子法），主要参考内容是原文的补充材料，关于论文其他部分创新点及其整体思路会在后续文章中进行分析。（笔记1的链接：http://blog.csdn.net/discoverer100/article/details/78182306）

2. 滤波器求解目标函数的构建

在多通道情况下，目标函数为

argminh∑d=1Nd(∥fd⊙hd−g∥2+λ∥hd∥2)=argminh∑d=1Nd(∥∥h^Hddiag(f^d)−g^d∥∥2+λ∥∥h^d∥∥2)(1)$$\begin{array}{}\text{(1)}& \begin{array}{c}\underset{\mathbf{h}}{\arg min}\sum _{d=1}^{N_d}\left({\Vert {\mathbf{f}}_d\odot {\mathbf{h}}_d-\mathbf{g}\Vert }^2+\lambda {\Vert {\mathbf{h}}_d\Vert }^2\right)\\ =\underset{\mathbf{h}}{\arg min}\sum _{d=1}^{N_d}\left({\Vert {\hat{\mathbf{h}}}_d^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left({\hat{\mathbf{f}}}_d\right)-{\hat{g}}_d\Vert }^2+\lambda {\Vert {\hat{\mathbf{h}}}_d\Vert }^2\right)\end{array}\end{array}$$

其中，h$\mathbf{h}$ 表示滤波器，d=1toNd$d=1to{N}_d$ 表示Nd$N_d$ 个通道，g$\mathbf{g}$ 表示期望的响应输出，λ$\lambda$ 表示正则项用于防止过拟合（关于正则项为什么可以防止过拟合可以参考：http://www.cnblogs.com/alexanderkun/p/6922428.html）

根据上述(1)式，为简化推导过程，将多通道情况改为单通道情况模式，则目标函数为

argminh∥f⊙h−g∥2+λ∥h∥2=argminh∥∥h^Hdiag(f^)−g^∥∥2+λ∥∥h^∥∥2(2)$$\begin{array}{}\text{(2)}& \begin{array}{c}\underset{\mathbf{h}}{\arg min}{\Vert \mathbf{f}\odot \mathbf{h}-\mathbf{g}\Vert }^2+\lambda {\Vert \mathbf{h}\Vert }^2\\ =\underset{\mathbf{h}}{\arg min}{\Vert {\hat{\mathbf{h}}}^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2+\lambda {\Vert \hat{\mathbf{h}}\Vert }^2\end{array}\end{array}$$

引入变量hc${\mathbf{h}}_{\mathbf{c}}$ 并定义约束条件

hc−hm=0(3)$$\begin{array}{}\text{(3)}& {\mathbf{h}}_c-{\mathbf{h}}_m=0\end{array}$$

其中，hm≡m⊙h${\mathbf{h}}_m\equiv \mathbf{m}\odot \mathbf{h}$ ，而m$\mathbf{m}$ 表示论文中的空间置信图（spatial reliability map），也可以理解为一个mask，具体概念可以参考前面的一篇文章：http://blog.csdn.net/discoverer100/article/details/78182306，上述(3)式中引入的变量hc${\mathbf{h}}_{\mathbf{c}}$ 可以先不理会其物理意义，它的主要作用是让算法能够收敛（论文原文表述：prohibits a closed-form solution），个人猜测：这里的下标命名为c，可能就是取constrained的第一个字母。

对(2)式引入上述约束条件，并进一步调整，得到最终的目标函数

argminhc,hm∥∥h^Hcdiag(f^)−g^∥∥2+λ2∥∥h^m∥∥2s.t.hc−hm=0(4)$$\begin{array}{}\text{(4)}& \begin{array}{l}\underset{{\mathbf{h}}_c,{\mathbf{h}}_m}{\arg min}{\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2+\frac{\lambda }{2}{\Vert {\hat{\mathbf{h}}}_m\Vert }^2\\ s.t.{\mathbf{h}}_c-{\mathbf{h}}_m=0\end{array}\end{array}$$

上述的正则项前面多出了一个系数1/2$1/2$ ，其主要意图是求导数后系数可以变为1$1$ ，便于公式书写。

这样，公式(4)就是我们推导的起始表达式。

3. 构建Lagrange表达式

根据上述目标函数，以及Augmented Lagrangian方法（参考Distributed optimization and statistical learning via the alternating direction method of multipliers），构建Lagrang表达式，如下

L(h^c,h,I^|m)=∥∥h^Hcdiag(f^)−g^∥∥2+λ2∥hm∥2+[I^H(h^c−h^m)+I^H(h^c−h^m)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]+μ∥∥h^c−h^m∥∥2(5)$$\begin{array}{}\text{(5)}& \mathcal{L}\left({\hat{\mathbf{h}}}_c,\mathbf{h},\hat{\mathbf{I}}|\mathbf{m}\right)={\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2+\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2+\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-{\hat{\mathbf{h}}}_m\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-{\hat{\mathbf{h}}}_m\right)}\right]+\mu {\Vert {\hat{\mathbf{h}}}_c-{\hat{\mathbf{h}}}_m\Vert }^2\end{array}$$

其中，字母I$\mathbf{I}$ 表示Lagrange乘数，字母上面的横杠表示共轭矩阵，字母右上方的H$H$ 表示共轭转置矩阵，因此有规律：A¯T=AH${\overline{\mathbf{A}}}^T={\mathbf{A}}^H$ （后面的推导中可能同时存在两种表示，需要留意）。将上述(5)式进行向量化表示，可得

L(h^c,h,I^|m)=∥∥h^Hcdiag(f^)−g^∥∥2+λ2∥hm∥2+[I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]+μ∥∥h^c−D−−√FMh∥∥2(6)$$\begin{array}{}\text{(6)}& \mathcal{L}\left({\hat{\mathbf{h}}}_c,\mathbf{h},\hat{\mathbf{I}}|\mathbf{m}\right)={\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2+\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2+\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]+\mu {\Vert {\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\Vert }^2\end{array}$$

不难看出，上述(5)式到(6)式，主要变化就是将变量h^m${\hat{\mathbf{h}}}_m$ 的表达式替换为D−−√FMh$\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}$ ，其中F$F$ 表示离散傅里叶变换矩阵，它相当于一个常量，D$D$ 是F$F$ 的大小（F$F$ 是一个D×D$D\times D$ 的方阵），M=diag(m)$\mathbf{M}\mathbf{=}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\mathbf{m}\right)$

将上述(6)式简单表述为四个项的和，为

L(h^c,h,I^)=L1+L2+L3+L4(7)$$\begin{array}{}\text{(7)}& \mathcal{L}\left({\hat{\mathbf{h}}}_c,\mathbf{h},\hat{\mathbf{I}}\right)={\mathcal{L}}_1+{\mathcal{L}}_2+{\mathcal{L}}_3+{\mathcal{L}}_4\end{array}$$

其中，

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪L1=∥∥h^Hcdiag(f^)−g^∥∥2=(h^Hcdiag(f^)−g^)(h^Hcdiag(f^)−g^)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯TL2=λ2∥hm∥2L3=I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯L4=μ∥∥h^c−D−−√FMh∥∥2=μ(h^c−D−−√FMh)(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T(8)$$\begin{array}{}\text{(8)}& \{\begin{array}{l}{\mathcal{L}}_1={\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2=\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right){\overline{\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right)}}^T\\ {\mathcal{L}}_2=\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2\\ {\mathcal{L}}_3={\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\\ {\mathcal{L}}_4=\mu {\Vert {\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\Vert }^2=\mu \left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\overline{\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^T\end{array}\end{array}$$

4. 开始优化，首先对h_c求偏导数

对上述公式(4)的优化可以表述为下面的迭代过程

h^optc=argminhcL(h^c,h,I^)hopt=argminhL(h^optc,h,I^)(9)$$\begin{array}{}\text{(9)}& \begin{array}{l}{\hat{\mathbf{h}}}_c^{\mathrm{o}\mathrm{p}\mathrm{t}}=\underset{{\mathbf{h}}_c}{\arg min}L\left({\hat{\mathbf{h}}}_c,\mathbf{h},\hat{\mathbf{I}}\right)\\ {\mathbf{h}}^{\mathrm{o}\mathrm{p}\mathrm{t}}=\underset{\mathbf{h}}{\arg min}L\left({\hat{\mathbf{h}}}_c^{\mathrm{o}\mathrm{p}\mathrm{t}},\mathbf{h},\hat{\mathbf{I}}\right)\end{array}\end{array}$$

现在看关于变量h^c${\hat{\mathbf{h}}}_c$ 的优化，需要令满足∇h^c¯¯¯¯¯¯L≡0${\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}\mathcal{L}\equiv 0$ ，也就是

∇h^c¯¯¯¯¯¯L1+∇h^c¯¯¯¯¯¯L2+∇h^c¯¯¯¯¯¯L3+∇h^c¯¯¯¯¯¯L4≡0(10)$$\begin{array}{}\text{(10)}& {\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_1+{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_2+{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_3+{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_4\equiv 0\end{array}$$

对各个分量求偏导数，有

∇h^c¯¯¯¯¯¯L1=∂∂h^c¯¯¯¯¯[(h^Hcdiag(f^)−g^)(h^Hcdiag(f^)−g^)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T]=∂∂h^c¯¯¯¯¯[h^Hcdiag(f^)diag(f^)Hh^c−h^Hcdiag(f^)g^H−g^diag(f^)Hh^c+g^g^H]=diag(f^)diag(f^)Hh^c−diag(f^)g^H−0+0=diag(f^)diag(f^)Hh^c−diag(f^)g^H(11)$$\begin{array}{}\text{(11)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_1& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right){\overline{\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right)}}^T\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[{\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H-\hat{\mathbf{g}}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c+\hat{\mathbf{g}}{\hat{\mathbf{g}}}^H\right]\\ & =\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H-0+0\\ & =\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H\end{array}\end{array}$$

∇h^c¯¯¯¯¯¯L2=∇h^c¯¯¯¯¯¯[λ2∥hm∥2]=0(12)$$\begin{array}{}\text{(12)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_2& ={\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}\left[\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2\right]\\ & =0\end{array}\end{array}$$

∇h^c¯¯¯¯¯¯L3=∂∂h^c¯¯¯¯¯[I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h^c¯¯¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^TD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=0−0+I^T−0=I^(13)$$\begin{array}{}\text{(13)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_3& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-{\hat{\mathbf{I}}}^T\overline{\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =0-0+{\hat{\mathbf{I}}}^T-0\\ & =\hat{\mathbf{I}}\end{array}\end{array}$$

∇h^c¯¯¯¯¯¯L4=∂∂h^c¯¯¯¯¯[μ(h^c−D−−√FMh)(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T]=∂∂h^c¯¯¯¯¯[μ(h^c⋅h^Hc−h^c⋅D−−√hHMFH−D−−√FMhh^Hc+D−−√FMh⋅D−−√hHMFH)]=∂∂h^c¯¯¯¯¯[μ(h^c⋅h^Hc−h^c⋅D−−√hHMFH−D−−√FMhh^Hc+DFMhhHMFH)]=μ(h^c−0−D−−√FMh+0)=μh^c−μD−−√FMh(14)$$\begin{array}{}\text{(14)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{{\hat{\mathbf{h}}}_c}}{\mathcal{L}}_4& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[\mu \left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\overline{\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^T\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[\mu \left({\hat{\mathbf{h}}}_c\cdot {\hat{\mathbf{h}}}_c^H-{\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\hat{\mathbf{h}}}_c^H+\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right)\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{{\hat{\mathbf{h}}}_c}}\left[\mu \left({\hat{\mathbf{h}}}_c\cdot {\hat{\mathbf{h}}}_c^H-{\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\hat{\mathbf{h}}}_c^H+D\mathbf{F}\mathbf{M}\mathbf{h}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right)\right]\\ & =\mu \left({\hat{\mathbf{h}}}_c-0-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+0\right)\\ & =\mu {\hat{\mathbf{h}}}_c-\mu \sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\end{array}\end{array}$$

于是，上述公式(10)可以写成

diag(f^)diag(f^)Hh^c−diag(f^)g^H+0+I^+μ(h^c−0−D−−√FMh+0)≡0diag(f^)diag(f^)Hh^c−diag(f^)g^H+I^+μ(h^c−0−D−−√FMh+0)≡0(15)$$\begin{array}{}\text{(15)}& \begin{array}{c}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H+0+\hat{\mathbf{I}}+\mu \left({\hat{\mathbf{h}}}_c-0-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+0\right)\equiv 0\\ \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H+\hat{\mathbf{I}}+\mu \left({\hat{\mathbf{h}}}_c-0-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+0\right)\equiv 0\end{array}\end{array}$$

回顾公式(6)，我们曾将变量h^m${\hat{\mathbf{h}}}_m$ 的表达式替换为D−−√FMh$\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}$ ，现在我们将它替换回来，得

diag(f^)diag(f^)Hh^c−diag(f^)g^H+I^+μ(h^c−0−h^m+0)≡0(16)$$\begin{array}{}\text{(16)}& \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H{\hat{\mathbf{h}}}_c-\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H+\hat{\mathbf{I}}+\mu \left({\hat{\mathbf{h}}}_c-0-{\hat{\mathbf{h}}}_m+0\right)\equiv 0\end{array}$$

针对h^c${\hat{\mathbf{h}}}_c$ 合并同类项，得

h^c⋅[diag(f^)diag(f^)H+μ]=μh^m+diag(f^)g^H−I^(17)$$\begin{array}{}\text{(17)}& {\hat{\mathbf{h}}}_c\cdot \left[\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H+\mu \right]=\mu {\hat{\mathbf{h}}}_m+\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H-\hat{\mathbf{I}}\end{array}$$

于是，

h^c=diag(f^)g^H+μh^m−I^diag(f^)diag(f^)H+μ(18)$$\begin{array}{}\text{(18)}& {\hat{\mathbf{h}}}_c=\frac{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right){\hat{\mathbf{g}}}^H+\mu {\hat{\mathbf{h}}}_m-\hat{\mathbf{I}}}{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}{\left(\hat{\mathbf{f}}\right)}^H+\mu }\end{array}$$

根据对角矩阵的乘法性质以及相关滤波基础概念，我们可以将上述(18)式的表达进行简化，得

h^c=f^⊙g^∗+μh^m−I^f^⊙f^∗+μ(19)$$\begin{array}{}\text{(19)}& {\hat{\mathbf{h}}}_c=\frac{\hat{\mathbf{f}}\odot {\hat{\mathbf{g}}}^{\ast }+\mu {\hat{\mathbf{h}}}_m-\hat{\mathbf{I}}}{\hat{\mathbf{f}}\odot {\hat{\mathbf{f}}}^{\ast }+\mu }\end{array}$$

由于右上角得星号表示共轭矩阵，为与作者原文表述一致，也可用顶部的横杠表示，有

h^c=f^⊙g^¯¯¯+μh^m−I^f^⊙f^¯¯¯+μ(20)$$\begin{array}{}\text{(20)}& {\hat{\mathbf{h}}}_c=\frac{\hat{\mathbf{f}}\odot \overline{\hat{\mathbf{g}}}+\mu {\hat{\mathbf{h}}}_m-\hat{\mathbf{I}}}{\hat{\mathbf{f}}\odot \overline{\hat{\mathbf{f}}}+\mu }\end{array}$$

这样就完成了对变量h^c${\hat{\mathbf{h}}}_c$ 的最优化求解，它对应论文官方源码中的变量G（位于create_csr_filter.m中）

G = (Sxy + mu*H - L) ./ (Sxx + mu);

5. 对变量h求解偏导数

前面的公式(10)-(20)都是针对变量h^c${\hat{\mathbf{h}}}_c$ 求解偏导数，由于论文提出的Lagrange表达式中含有两个变量，现在还需要针对变量h¯¯¯$\overline{\mathbf{h}}$ 求解偏导数，也就是

∇h¯¯¯¯L1+∇h¯¯¯¯L2+∇h¯¯¯¯L3+∇h¯¯¯¯L4≡0(21)$$\begin{array}{}\text{(21)}& {\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_1+{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_2+{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_3+{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_4\equiv 0\end{array}$$

首先回顾(8)式，也就是L1、L2、L3和L4这四个优化项

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪L1=∥∥h^Hcdiag(f^)−g^∥∥2=(h^Hcdiag(f^)−g^)(h^Hcdiag(f^)−g^)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯TL2=λ2∥hm∥2L3=I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯L4=μ∥∥h^c−D−−√FMh∥∥2=μ(h^c−D−−√FMh)(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T(8)$$\begin{array}{}\text{(8)}& \{\begin{array}{l}{\mathcal{L}}_1={\Vert {\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{g}\Vert }^2=\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right){\overline{\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right)}}^T\\ {\mathcal{L}}_2=\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2\\ {\mathcal{L}}_3={\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\\ {\mathcal{L}}_4=\mu {\Vert {\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\Vert }^2=\mu \left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\overline{\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^T\end{array}\end{array}$$

其关于变量h¯¯¯$\overline{\mathbf{h}}$ 的偏导数为

∇h¯¯¯¯L1=∂∂h¯¯¯[(h^Hcdiag(f^)−g^)(h^Hcdiag(f^)−g^)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T]=0(22)$$\begin{array}{}\text{(22)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_1& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right){\overline{\left({\hat{\mathbf{h}}}_c^H\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\hat{\mathbf{f}}\right)-\hat{\mathbf{g}}\right)}}^T\right]\\ & =0\end{array}\end{array}$$

∇h¯¯¯¯L2=∂∂h¯¯¯[λ2∥hm∥2]=∂∂h¯¯¯[λ2∥m⊙h∥2]=∂∂h¯¯¯[λ2(Mh)¯¯¯¯¯¯¯¯¯¯¯¯T(Mh)]=∂∂h¯¯¯[λ2hHMHMh](23)$$\begin{array}{}\text{(23)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_2& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\Vert {\mathbf{h}}_m\Vert }^2\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\Vert \mathbf{m}\odot \mathbf{h}\Vert }^2\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\overline{\left(\mathbf{M}\mathbf{h}\right)}}^T\left(\mathbf{M}\mathbf{h}\right)\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\mathbf{h}}^H{\mathbf{M}}^H\mathbf{M}\mathbf{h}\right]\end{array}\end{array}$$

观察(23)式，由于其中的M$\mathbf{M}$ 表示论文定义的mask（也就是Spatial reliability map）的对角线元素，这个矩阵中的所有元素仅为0或者1，它们都是实数，因此(23)式中的MHM${\mathbf{M}}^H\mathbf{M}$ 可以简化表示为M$\mathbf{M}$ ，于是，

∇h¯¯¯¯L2=∂∂h¯¯¯[λ2hHMh]=∂∂h¯¯¯[λ2((hHMh)T)T]=∂∂h¯¯¯[λ2(hTMTh¯¯¯)T]=(λ2(hTMT))T=λ2Mh(24)$$\begin{array}{}\text{(24)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_2& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\mathbf{h}}^H\mathbf{M}\mathbf{h}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\left({\left({\mathbf{h}}^H\mathbf{M}\mathbf{h}\right)}^T\right)}^T\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\left({\mathbf{h}}^T{\mathbf{M}}^T\overline{\mathbf{h}}\right)}^T\right]\\ & ={\left(\frac{\lambda }{2}\left({\mathbf{h}}^T{\mathbf{M}}^T\right)\right)}^T\\ & =\frac{\lambda }{2}\mathbf{M}\mathbf{h}\end{array}\end{array}$$

2018年8月更新：特别注意，上述公式(24)推导过程有误，应改为如下（此处特别感谢CSDN网友weixin_41660496）：

c∇h¯¯¯¯L2=∂∂h¯¯¯[λ2hHMh]=∂∂h¯¯¯[λ2(h¯¯¯)TMh]=λ2Mh(24)$$\begin{array}{}\text{(24)}& \begin{array}{rl}c{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{L}_2& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\mathbf{h}}^H\mathbf{M}\mathbf{h}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\frac{\lambda }{2}{\left(\overline{\mathbf{h}}\right)}^T\mathbf{M}\mathbf{h}\right]\\ & =\frac{\lambda }{2}\mathbf{M}\mathbf{h}\end{array}\end{array}$$

∇h¯¯¯¯L3=∂∂h¯¯¯[I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^TD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=0−0+0−I^TD−−√FM¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯=−D−−√⋅I^T⋅F¯¯¯¯⋅M¯¯¯¯¯¯=−D−−√⋅(I^)T⋅(FH)T⋅(MH)T=−D−−√(MHFHI^)T=−D−−√(MFHI^)T(25)$$\begin{array}{}\text{(25)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_3& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-{\hat{\mathbf{I}}}^T\overline{\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =0-0+0-{\hat{\mathbf{I}}}^T\overline{\sqrt{D}\mathbf{F}\mathbf{M}}\\ & =-\sqrt{D}\cdot {\hat{\mathbf{I}}}^T\cdot \overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\\ & =-\sqrt{D}\cdot {\left(\hat{\mathbf{I}}\right)}^T\cdot {\left({\mathbf{F}}^H\right)}^T\cdot {\left({\mathbf{M}}^H\right)}^T\\ & =-\sqrt{D}{\left({\mathbf{M}}^H{\mathbf{F}}^H\hat{\mathbf{I}}\right)}^T\\ & =-\sqrt{D}{\left(\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}\right)}^T\end{array}\end{array}$$

2018年8月更新：特别注意，上述公式(25)推导过程有误，应改为如下（此处特别感谢CSDN网友weixin_41660496）：

∇h¯¯¯¯L3=∂∂h¯¯¯[I^H(h^c−D−−√FMh)+I^H(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^H(h^c−D−−√FMh)+I^Hh^c¯¯¯¯¯¯¯¯¯¯¯−I^HD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^HD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^HD−−√FMh¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^H¯¯¯¯¯¯⋅D−−√¯¯¯¯¯¯¯¯¯⋅FMh¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^H¯¯¯¯¯¯⋅D−−√⋅FMh¯¯¯¯¯¯¯¯¯¯¯¯]=∂∂h¯¯¯[I^Hh^c−I^HD−−√FMh+I^Th^c¯¯¯¯¯−I^H¯¯¯¯¯¯⋅D−−√⋅F¯¯¯¯⋅M¯¯¯¯¯¯⋅h¯¯¯]=0−0+0−(I^H¯¯¯¯¯¯⋅D−−√)⋅(F¯¯¯¯⋅M¯¯¯¯¯¯)T=0−0+0−(I^H¯¯¯¯¯¯⋅D−−√)⋅M¯¯¯¯¯¯T⋅F¯¯¯¯T=0−0+0−(I^⋅D−−√)⋅MT⋅FH=−D−−√MTFHI^=−D−−√MFHI^$$\begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{L}_3& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)+\overline{{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H}\cdot \overline{\sqrt{D}}\cdot \overline{\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H}\cdot \sqrt{D}\cdot \overline{\mathbf{F}\mathbf{M}\mathbf{h}}\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[{\hat{\mathbf{I}}}^H{\hat{\mathbf{h}}}_c-{\hat{\mathbf{I}}}^H\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}+{\hat{\mathbf{I}}}^T\overline{{\hat{\mathbf{h}}}_c}-\overline{{\hat{\mathbf{I}}}^H}\cdot \sqrt{D}\cdot \overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\cdot \overline{\mathbf{h}}\right]\\ & =0-0+0-\left(\overline{{\hat{\mathbf{I}}}^H}\cdot \sqrt{D}\right)\cdot {\left(\overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\right)}^T\\ & =0-0+0-\left(\overline{{\hat{\mathbf{I}}}^H}\cdot \sqrt{D}\right)\cdot {\overline{\mathbf{M}}}^T\cdot {\overline{\mathbf{F}}}^T\\ & =0-0+0-\left(\hat{\mathbf{I}}\cdot \sqrt{D}\right)\cdot {\mathbf{M}}^T\cdot {\mathbf{F}}^H\\ & =-\sqrt{D}{\mathbf{M}}^T{\mathbf{F}}^H\hat{\mathbf{I}}\\ & =-\sqrt{D}\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}\end{array}$$

上述(25)式中，利用了矩阵D−−√,M$\sqrt{D},\mathbf{M}$ 为实数矩阵的性质，所以有M¯¯¯¯¯¯=M$\overline{\mathbf{M}}=\mathbf{M}$ , D¯¯¯¯¯=D$\overline{\mathbf{D}}=\mathbf{D}$ ，另外M$\mathbf{M}$ 也是一个对角矩阵，因此MT=M${\mathbf{M}}^T=\mathbf{M}$ 。

最后是L4项

∇h¯¯¯¯L4=∂∂h¯¯¯[μ(h^c−D−−√FMh)(h^c−D−−√FMh)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯T]=∂∂h¯¯¯[μ(h^c⋅h^Hc−h^c⋅D−−√(F¯¯¯¯⋅M¯¯¯¯¯¯⋅h¯¯¯)T−D−−√FMhh^Hc+DFMh(F¯¯¯¯⋅M¯¯¯¯¯¯⋅h¯¯¯)T)]=∂∂h¯¯¯[μ(h^c⋅h^Hc−h^c⋅D−−√hHMFH−D−−√FMhh^Hc+DFMhhHMHFH)]=0−∂∂h¯¯¯[μ⋅h^c⋅D−−√hHMFH]−0+∂∂h¯¯¯[μ⋅DFMhhHMHFH]=−[∂∂hH[μ⋅h^c⋅D−−√hHMFH]]T+[∂∂hH[μ⋅DFMhhHMHFH]]T=−[∂∂hH[μ⋅D−−√⋅h^c⋅hHMFH]]T+[∂∂hH[μ⋅DFMh⋅hH⋅MHFH]]T=−μ⋅D−−√⋅[(h^c)T⋅(MFH)T]T+μ⋅D⋅[(FMh)T⋅(MHFH)T]T=−μ⋅D−−√⋅[(MFH)T]T⋅[(h^c)T]T+μ⋅D⋅[(MHFH)T]T⋅[(FMh)T]T=−μ⋅D−−√⋅MFH⋅h^c+μ⋅D⋅MHFH⋅FMh=−μ⋅D−−√⋅MFH⋅h^c+μ⋅D⋅MHIMh=−μ⋅D−−√⋅MFH⋅h^c+μ⋅D⋅(M⋅M)h=−μD−−√MFHh^c+μDMh(26)$$\begin{array}{}\text{(26)}& \begin{array}{rl}{\mathrm{\nabla }}_{\overline{\mathbf{h}}}{\mathcal{L}}_4& =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right){\overline{\left({\hat{\mathbf{h}}}_c-\sqrt{D}\mathbf{F}\mathbf{M}\mathbf{h}\right)}}^T\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \left({\hat{\mathbf{h}}}_c\cdot {\hat{\mathbf{h}}}_c^H-{\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\left(\overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\cdot \overline{\mathbf{h}}\right)}^T-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\hat{\mathbf{h}}}_c^H+D\mathbf{F}\mathbf{M}\mathbf{h}{\left(\overline{\mathbf{F}}\cdot \overline{\mathbf{M}}\cdot \overline{\mathbf{h}}\right)}^T\right)\right]\\ & =\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \left({\hat{\mathbf{h}}}_c\cdot {\hat{\mathbf{h}}}_c^H-{\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H-\sqrt{D}{\mathbf{F}\mathbf{M}\mathbf{h}\hat{\mathbf{h}}}_c^H+D\mathbf{F}\mathbf{M}\mathbf{h}{\mathbf{h}}^H{\mathbf{M}}^H{\mathbf{F}}^H\right)\right]\\ & =0-\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \cdot {\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right]-0+\frac{\mathrm{\partial }}{\mathrm{\partial }\overline{\mathbf{h}}}\left[\mu \cdot D\mathbf{F}\mathbf{M}\mathbf{h}{\mathbf{h}}^H{\mathbf{M}}^H{\mathbf{F}}^H\right]\\ & =-{\left[\frac{\mathrm{\partial }}{\mathrm{\partial }{\mathbf{h}}^H}\left[\mu \cdot {\hat{\mathbf{h}}}_c\cdot \sqrt{D}{\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right]\right]}^T+{\left[\frac{\mathrm{\partial }}{\mathrm{\partial }{\mathbf{h}}^H}\left[\mu \cdot D\mathbf{F}\mathbf{M}\mathbf{h}{\mathbf{h}}^H{\mathbf{M}}^H{\mathbf{F}}^H\right]\right]}^T\\ & =-{\left[\frac{\mathrm{\partial }}{\mathrm{\partial }{\mathbf{h}}^H}\left[\mu \cdot \sqrt{D}\cdot {\hat{\mathbf{h}}}_c\cdot {\mathbf{h}}^H\mathbf{M}{\mathbf{F}}^H\right]\right]}^T+{\left[\frac{\mathrm{\partial }}{\mathrm{\partial }{\mathbf{h}}^H}\left[\mu \cdot D\mathbf{F}\mathbf{M}\mathbf{h}\cdot {\mathbf{h}}^H\cdot {\mathbf{M}}^H{\mathbf{F}}^H\right]\right]}^T\\ & =-\mu \cdot \sqrt{D}\cdot {\left[{\left({\hat{\mathbf{h}}}_c\right)}^T\cdot {\left(\mathbf{M}{\mathbf{F}}^H\right)}^T\right]}^T+\mu \cdot D\cdot {\left[{\left(\mathbf{F}\mathbf{M}\mathbf{h}\right)}^T\cdot {\left({\mathbf{M}}^H{\mathbf{F}}^H\right)}^T\right]}^T\\ & =-\mu \cdot \sqrt{D}\cdot {\left[{\left(\mathbf{M}{\mathbf{F}}^H\right)}^T\right]}^T\cdot {\left[{\left({\hat{\mathbf{h}}}_c\right)}^T\right]}^T+\mu \cdot D\cdot {\left[{\left({\mathbf{M}}^H{\mathbf{F}}^H\right)}^T\right]}^T\cdot {\left[{\left(\mathbf{F}\mathbf{M}\mathbf{h}\right)}^T\right]}^T\\ & =-\mu \cdot \sqrt{D}\cdot \mathbf{M}{\mathbf{F}}^H\cdot {\hat{\mathbf{h}}}_c+\mu \cdot D\cdot {\mathbf{M}}^H{\mathbf{F}}^H\cdot \mathbf{F}\mathbf{M}\mathbf{h}\\ & =-\mu \cdot \sqrt{D}\cdot \mathbf{M}{\mathbf{F}}^H\cdot {\hat{\mathbf{h}}}_c+\mu \cdot D\cdot {\mathbf{M}}^H\mathbf{I}\mathbf{M}\mathbf{h}\\ & =-\mu \cdot \sqrt{D}\cdot \mathbf{M}{\mathbf{F}}^H\cdot {\hat{\mathbf{h}}}_c+\mu \cdot D\cdot \left(\mathbf{M}\cdot \mathbf{M}\right)\mathbf{h}\\ & =-\mu \sqrt{D}\mathbf{M}{\mathbf{F}}^H{\hat{\mathbf{h}}}_c+\mu D\mathbf{M}\mathbf{h}\end{array}\end{array}$$

上述(26)式末尾利用了MM=M$\mathbf{M}\mathbf{M}=\mathbf{M}$ ，这是因为M$\mathbf{M}$ 表示空间置信图（Spatial reliability map），根据论文中的定义，矩阵M$\mathbf{M}$ 中的元素要么为1，要么为0，因此MM=M$\mathbf{M}\mathbf{M}=\mathbf{M}$

将上述(22)-(26)式代入到(21)式中，有

λ2Mh−D−−√MFHI^−μD−−√MFHh^c+μDMh=0(27)$$\begin{array}{}\text{(27)}& \frac{\lambda }{2}\mathbf{M}\mathbf{h}-\sqrt{D}\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}-\mu \sqrt{D}\mathbf{M}{\mathbf{F}}^H{\hat{\mathbf{h}}}_c+\mu D\mathbf{M}\mathbf{h}=0\end{array}$$

对上式中的Mh$\mathbf{M}\mathbf{h}$ 进行合并同类项，有

(λ2+μD)MhMhMhMhMh=D−−√MFHI^+μD−−√MFHh^c=D−−√MFHI^+μD−−√MFHh^cλ2+μD=M⋅D−−√FH(I^+μh^c)λ2+μD=M⋅D√DFH(I^+μh^c)1D(λ2+μD)=M⋅1D√FH(I^+μh^c)λ2D+μ(28)$$\begin{array}{}\text{(28)}& \begin{array}{rl}\left(\frac{\lambda }{2}+\mu D\right)\mathbf{M}\mathbf{h}& =\sqrt{D}\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}+\mu \sqrt{D}\mathbf{M}{\mathbf{F}}^H{\hat{\mathbf{h}}}_c\\ \mathbf{M}\mathbf{h}& =\frac{\sqrt{D}\mathbf{M}{\mathbf{F}}^H\hat{\mathbf{I}}+\mu \sqrt{D}\mathbf{M}{\mathbf{F}}^H{\hat{\mathbf{h}}}_c}{\frac{\lambda }{2}+\mu D}\\ \mathbf{M}\mathbf{h}& =\mathbf{M}\cdot \frac{\sqrt{D}{\mathbf{F}}^H\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2}+\mu D}\\ \mathbf{M}\mathbf{h}& =\mathbf{M}\cdot \frac{\frac{\sqrt{D}}{D}{\mathbf{F}}^H\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{1}{D}\left(\frac{\lambda }{2}+\mu D\right)}\\ \mathbf{M}\mathbf{h}& =\mathbf{M}\cdot \frac{\frac{1}{\sqrt{D}}{\mathbf{F}}^H\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\end{array}\end{array}$$

现在简单回顾一下傅里叶逆变换的定义

F−1(x^)=1D−−√FHx^(29)$$\begin{array}{}\text{(29)}& {\mathcal{F}}^{-1}\left(\hat{\mathbf{x}}\right)=\frac{1}{\sqrt{D}}{\mathbf{F}}^H\hat{\mathbf{x}}\end{array}$$

据此，可以将上述(28)式表示为

Mh=M⋅F−1(I^+μh^c)λ2D+μdiag(m)h=diag(m)⋅F−1(I^+μh^c)λ2D+μm⊙h=m⊙F−1(I^+μh^c)λ2D+μ(30)$$\begin{array}{}\text{(30)}& \begin{array}{r}\mathbf{M}\mathbf{h}=\mathbf{M}\cdot \frac{{\mathcal{F}}^{-1}\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\\ \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\mathbf{m}\right)\mathbf{h}=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\mathbf{m}\right)\cdot \frac{{\mathcal{F}}^{-1}\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\\ \mathbf{m}\odot \mathbf{h}=\mathbf{m}\odot \frac{{\mathcal{F}}^{-1}\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\end{array}\end{array}$$

由于论文中最优化表达式的约束为m≡m⊙h$\mathbf{m}\equiv \mathbf{m}\odot \mathbf{h}$ （这里左边的h$\mathbf{h}$ 实际上也就是hm${\mathbf{h}}_m$ ），因此上述(30)式也可以表示为

h=m⊙F−1(I^+μh^c)λ2D+μ(31)$$\begin{array}{}\text{(31)}& \mathbf{h}=\mathbf{m}\odot \frac{{\mathcal{F}}^{-1}\left(\hat{\mathbf{I}}+\mu {\hat{\mathbf{h}}}_c\right)}{\frac{\lambda }{2D}+\mu }\end{array}$$

上述(31)式就对应论文原文中的公式(10)，相应的代码位于create_csr_filter.m中的变量H

H = fft2(real((1/(lambda + mu)) * bsxfun(@times, P, ifft2(mu*G + L))));

6. 小结

论文为求解最优的滤波器值，在原始目标函数的基础上，引入约束条件，利用Augmented Lagrangian方法构造最优化表达式，最后利用偏导数值为0的情况进行求解，这种利用拉格朗日最优化方法进行目标跟踪算法的建模思想值得我们学习，关于(25)式的转置问题，也欢迎大家共同讨论交流。

---

更多内容，请扫码关注“视觉边疆”微信订阅号